a 



72-6/3 



Conservation Resources 
Lig-Free® Type I 
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The Calculation of 
End Areas 



BY 



E. S. M. LOVELACE, B.A.Sc. 

M. Can. Soc. C. E. 
Civil and Consulting Engineer 



* 



Copyright, 1914, by Edgar S. M. Lovelace 



4 Hospital Street, 
Montreal 



Copyrighted in Canada, 1914 



To be copyrighted in Canada, Great Britain and United States. 



A method of calculating end areas without any reduction of eleva- 
tions or plotting of cross sections, using direct the notes as taken in the 
field, thus effecting a great saving in time, giving increased accuracy, 
and at the same time reducing to a minimum the liability to error. 



GENERAL PROPOSITION 

Let P0P1P2P3P4P6, figures 1 and 2, be any polygonal figures and OX, 
OY any two lines at right angles to one another, and let the distances 
aia2a3. . . .represent the horizontal distances of the points P1P2P3. • . . 
from the vertical lines OY, which in the examples given coincide with 
the sides P0P5. 

Let h hih 2 h3 .... be the vertical distances of the points P0P1P2P3 .... 
from the lines OX giving the + sign to the distances below, and the — 
sign to those distances above OX. Then, the areas of the polygons 
P0P1P2P3P4P5 will be given by taking one half of the algebraic sum of the 
products found by multiplying consecutively each horizontal distance 
aia 2 a 3 ... .by the algebraic difference between the ordinates of the points 
coming immediately after and before such horizontal distance: that is 
area of the polygon P0P1P2P3P4P5 for figure 1 will be 
%\ ai(h 2 — ho)+a 2 (h3— hi)+a 3 (h4— h 2 )+a 4 (h 6 — h 3 ) \ 
and for figure 2 

Y 2 \ ai(h 2 — ho)+a2(— hs— hi)+a 3 (-h4-h2)+a 4 (— hs+hs) \ 
or y 2 \ ai(h 2 -ho)— a 2 (h 3 +hi)— a 3 (h 4 +h 2 ) -a 4 (h6 — h 3 ) \ 

The proof of the foregoing is easily seen, for drawing the horizontal 
dotted fines APi, BP 2 , CP 3 , DP4 and the diagonal dotted lines BPi, GP 2 , 
DP 3 , then for figure 1 

3^ai(h 2 — h ) = ^aiBPo = 3^a x (BA+AP ) = 3^aiBA+ ^aiAP =area 
BPoPx 

i^a 2 (h 3 -hi = 3^a 2 CA = ^a 2 (CB+BA) = ^a 2 CB + ^a 2 BA = area 
CBPxPa 

i^a 3 (h 4 -h 2 ) = ^a 3 DB = 3^a 3 (DC +CB) = ^a 3 DC + ^a 3 CB = area 
DCP 2 P 3 

3^a4(h 5 -h 3 ) = 3^a 4 P 5 C = Ha 4 (P 6 D +DC) = Ha 4 P 6 D + 3^a<DC = area 
P 6 DP 3 P4 and the sum of these equals the whole area P0P1P2P3P4P6. 

For figure 2 the proof is similar, for using the minus signs for the 
ordinates (corresponding to a fill or embankment) then 

i^ai(h 2 -h ) = ^aiBP = 3^ai(BA+AP ) = ^aiBA-f J^aiAPo = + 
area AP PiP 2 

- Ha 2 (h 3 +hi) = - ^a 2 (CA) = - i^a 2 (CB -BA) - - ^a 2 CB + 
%3iqBA= —area ACP 2 

APR -8 1914 
OCI.A371266 



^«fc. 




F»o. 2 



- ^a 3 (h 4 +h 2 ) = - y 2 2LzBT> = - ^a 3 (BC+CD) = - Ha 3 BC 
- Ha 3 CD = -area CP 2 P 3 D 

-H^Ou-ha) = -Ha 4 CP 5 = -Ha4(CD+DP 6 ) = -^a 4 CD 
-Ha 4 DP 6 = -area DP 3 P 4 P 6 
and the algebraic sum equals the whole area P PiP2P3P 4 P&. 

To adopt the above conclusions to field work so as to obtain direct 
from the level notes the areas of the cross sections as taken: — 

The horizontal line OX corresponds to the height of instrument and 
the line OY to a vertical line at centre. 

The ordinates hohih 2 h 3 are the rod readings taken at various 

points on the section and the horizontal distances aia»a s . . . .the cor- 
responding distances right or left of the C.L. 



tf> 



/0^ 



In the field book these are best noted in the form of a fraction as 
shown on the sample page, the numerator being the rod reading, the 
denominator the corresponding distance : 

The areas to the right and left of the centre line are calculated (as 
they are so taken in the field) separately and their sum gives the total 
area of the section. 

The horizontal distances to the left could of course be given the 
minus sign and exactly the same final result would be obtained, but in 
practice it is simpler to imagine the left half section as being drawn on 
the transparent left page of a book, so that when this page is turned 
over the left half section would be superimposed upon the right on the 
right hand page below, the centre line being represented by the line 
where the pages join. Thus all the horizontal distances, whether to 
right or left, can be given the plus sign. Taking the example at the 
head of the sample page, this section, if plotted, would appear as under. 




&'&&- Sl^P 



F«G 3 



By the formula 
area right = 3^ \ 6.0(2.4 -2.9) +13.0(1.9 -1.6) +20.4(8.8 -2.4) 
+10.0(8.8-1.9) \ 

= Y 2 \ -6.0,0.5+13.0,0.3+20.4,6.4+10.0,6.9) =100.23 

area left =Y 2 \ 8.0(7.1-2.9)+12.5(8.8-5.0)+10.0(8.8-7.1) \ 

= 1^(8.0,4.2 + 12.5,3.8+10.0,1.7) = 49.05 

/.Total End area 149.28 



In working out these areas in the field book the rod readings or 
ordinates can usually be subtracted mentally and their difference at 
once jotted down (with the proper sign + or — affixed), as shown on 
the marginal page. 

Should it be found impossible to complete the section with a single 
height of instrument, proceed as follows: suppose a section on side hill. 
(See P. 5) 

Then, if there be but one or two sections requiring a change in the 
height of instrument, it would only be necessary after setting up, to 
find on the upper side the difference 3.2 between 103.54 the new H.I. 
and 100.34 the original height, for then the rod readings as taken, sub- 
tracted from 3.2, would give the proper readings — 1.4 and —2.2 referred 
to the original H.I. 



&T. /as. s* 




28? 



Ft<5. A 



On the lower or left hand side of the section the new H.I. 93.92 being 
lower than 100.34, the original height, their difference 6.4 must be added 
to 2.4, the rod reading as taken, giving for this left side referred to the 
original H.I. a reading of 8.8. 

If, however, there are a number of consecutive sections requiring 
various changes as above, it would be better perhaps to enter on another 
page of the level book, under their respective H.I., the rod readings as 
taken, and then on the marginal page before using such for calculating 
the end areas reduce them as shown above to the original H.I. by adding 
or subtracting the constant difference. 

The original sections could then be completed by bringing back 
these corrected readings and the section for such a cross section as that 
illustrated above would appear in the level book as under. 



Station 


C. or F. 


B.S. 


H.I. 


F.S. 


Elevation 


Grade 


R 
40 

L 


+ 12.6 
+ 6.2 
+ 1.6 


4.2 



100.34 

3.1 1.6 
8.8 15.5 

5.6 
6.0 


See pa 
See pa 


ge. . —1.4 
21.6 

ge.. / 8.8 
V12.4 


r-2 2\ 

\ 28.9/ 
89.90 

) 



The resulting end area would then be as already shown: 
areaR = 3/2<i 8.8(1.6-4.2)+15.5(-1.4-3.1)+21.6(-2.2-1.6) 
+28.9(10.4 + 1.4) + 10.0(10.4+2.2) \ 
= Y 2 \ -8.8,2.6-15.5,4.5-21.6,3.8+28.9,11.8 + 10.0,12.6 \ =146.15 
area L=J^<j 6.0(8.8-4.2) + 12.4(10.4-5.6) + 10.0(10.4-8.8) \ 

= Y 2 \ 6.0,4.6 + 12.4,4.8 + 10.0,1.6 \ = 51.56 



.".Total End Area 197.71 



Sample page showing manner of keeping the field notes and the method 



Station 


C. or F. 


B.S. 


H.I. 


F.S. 


Elevation 


Grade 








100.34 








R 


+6.9 




1.6 2.4 
6.0 13.0 


/_L9 
V20.4 


) 




41 


+5.9 


2.9 









91.50 


L 


+ 1.7 




5.0 / 7.1\ 
8.0 V12.5/ 




















R 


+3.7 




5.5 / 5.3\ 
8.0 \15.5J 








41+40 


+2.9 


6.1 









91.30 


L 


+0.7 




6.7 / 8.3\ 
6.0 \11.0/ 




















41+62 


0.0 




Grade 






91.19 


R 


-1.6 




11.7 /10.9\ 
7.0 V10.4/ 








42 


-2.8 


12.1 









91.00 


L 


-5.4 




14.3 /14.7\ 
9.0 V16.1/ 




















T.P. 




0.84 


89.12 


12.06 


88.28 




R 


-3.0 




2.4 / 1.4\ 
9.0 V12.5/ 








42+60 


-4.6 


3.0 









90.70 


L 


-6.5 




3.1 4.9 
5.0 10.0 


/ 4.9 
V17.7 


) 





of calculating the end areas direct from these notes as taken. 



H.I. to 


Calculations, Remarks 


End Areas 


Grade 


ExcVn 


Emb'km't 




Road bed 20 ft. Exc. 16 ft. Bank 








-6.0,0.5+13.0,0.3+20.4,6.4+10.0,6.9 








2 




+8.8 




149.3 






+8.0,4.2 + 12.5,3.8 + 10.0, 1.7 
2 




% 




-8.0,0.8+15.5,3.5+10.0,3.7 
2 






+9.0 


+6.0,2.2 + 1 1.0,2.3 + 10.0,0.7 
2 


65.2 






• 


0.0 


00 




-7.0,1.2-10.4,2.4-8.0,1.6 
2 






+9.3 


+9.0,2.6-16.1,5.0-8.0,5.4 
2 

On rock 40 ft. Right of 42 

-9.0,1.6-12.5,4.0-8.0,3.0 
2 




73.2 


-1.6 






127.8 




+5.0, 1.9 + 10.0, 1.8 - 17.7,6.5 - 8.0,6.5 








2 





LIBRARY OF CONGRESS 



Station 



T.P. 



R 



43+50 



C. or F. 


B.S. 




9.16 


+4.0 




-0.4 


7.7 







-2.2 





H.I. 



T 021 607 849 4 

.1. _ . „».uu VJI1S 



89.12 
97.52 

Grade 

7^ ^2 
2.5 6.5 



8J. ( 9.5\ 

6.4 \n.3y 



0.76 



( 



3.3 
16.0 



88.36 



) 



90.25 



The section shown above is on side hill, embankment to the left, 
excavation to the right, the grade point being as shown 2.5 feet to the 
right. This must be taken into account in making the calculations so 
as to keep the excavation and embankment areas separate from one 
another. 

Thus, excavation area to the right is given by 
Y 2 \ 2.5(4.2-7.3)+6.5(3.3-7.3)+16.0(7.3-4.2)+10.0(7.3-3.3) \ 
= y 2 \ -2.5,3.1-6.5,4.0 + 16.0,3.1+10.0,4.0 \ =+27.9 

There would then be to the right in embankment the small area 
given by 2.5(7.3-7.7)3^ ' =-0.5 

and this would have to be added to the embankment area to the left 
given of course by 

Y 2 \ +6.4,1.8-11.3,0.8-8.0,2.2 |- = -7.6 that is a total area in em- 
bankment for the section of — 8.1 

It might however be considered less confusing in such cases to con- 
sider the vertical line as being placed at the grade point instead of at the 
centre of the section and then all the area to the right of this point would 
be in excavation and all to the left in embankment. 

This would simply mean that in applying the method all horizontal 
distances taken to the right would be reduced by 2.5 feet and all dis- 
tances to the left increased by the same amount, that is area right would 
= 3^ 4-0(3.3 -7.3) + 13.5(7.3 -4.2) +7.5(7.3 -3.3) \ 
= Y 2 \ -4.0,4.0+13.5,3.1+7.5,4.0 \ =in excavation 27.9 

as before, and area left would 

= y 2 \ 2.5(8.1-7.3)+8.9(9.5-7.7) + 13.8(7.3-8.1) + 10.5(7.3-9.5) \ 
= y 2 \ +2.5,0.8+8.9,1.8-13.8,0.8-10.5,2.2 \ equal as before to a 
total area in embankment of 8.1 



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